Answer:
6xy²+7-x²y²-5x²y+3x²y²=6x²+7+2x²y²-5x²y
X² + 1 = 0
=> (x+1)² - 2x = 0
=> x+1 = √(2x)
or x - √(2x) + 1 = 0
Now take y=√x
So, the equation changes to
y² - y√2 + 1 = 0
By quadratic formula, we get:-
y = [√2 ± √(2–4)]/2
or √x = (√2 ± i√2)/2 or (1 ± i)/√2 [by cancelling the √2 in numerator and denominator and ‘i' is a imaginary number with value √(-1)]
or x = [(1 ± i)²]/2
So roots are [(1+i)²]/2 and [(1 - i)²]/2
Thus we got two roots but in complex plane. If you put this values in the formula for formation of quadratic equation, that is x²+(a+b)x - ab where a and b are roots of the equation, you will get the equation
x² + 1 = 0 back again
So it’s x=1 or x=-1
Hi there!
To make n+1>4 true, you must have n be a numebr greater than 3...
The reason why is because if it is less than 3 or 3, it’ll have to be <.
The two numbers that can make this equation true is 4 and 5...
And if it is 1,2, and 3 it would be:
Are theese questions or your answers
if there answers they seem all right
